(a) Explanation Given information: The mass of the square plate is m . The side of a square plate is a . The angular velocity ( ω ) is ω 0 j Assume the impact to be perfectly plastic that is e = 0 . Calculation: Show the diagram of the system as in Figure (1). The length of the diagonal of a square is obtained by multiplying the side with 2 . The expression for the angular velocity in the x ′ axis ( ω x ′ ) as follows: ω x ′ = 2 2 ω 0 Here, ω 0 is the initial angular velocity. The expression for the angular velocity in the y ′ axis ( ω y ′ ) as follows: ω y ′ = 2 2 ω 0 The unit vectors along the x ′ and y ′ axis are represented by i ′ and j ′ . The expression for the initial angular momentum about the mass center ( H G ) 0 as follows: ( H G ) 0 = I ¯ x ′ ω x ′ i ′ + I ¯ y ′ ω y ′ j ′ (1) Here, ( H G ) 0 is the initial angular momentum about the mass center, I ¯ x ′ is the moment of inertia in the x ′ direction and I ¯ y ′ is the moment of inertia in the y ′ direction. The expression for the moment of inertia in the x ′ direction ( I ¯ x ′ ) as follows: I ¯ x ′ = 1 12 m a 2 Here, m is the mass and a is the side of the square plate. Due to symmetry, moment of inertia in the x ′ and y ′ axes are the same. The expression for the angular momentum about z ′ axis ( I ¯ z ′ ) as follows: I ¯ z ′ = 1 12 m a 2 Substitute 1 12 m a 2 for I ¯ x ′ , 1 12 m a 2 for I ¯ y ′ , 2 2 ω 0 for ω x ′ , and 2 2 ω 0 for ω y ′ in Equation (1). ( H G ) 0 = ( 1 12 m a 2 ) ( 2 2 ω 0 ) i ′ + ( 1 12 m a 2 ) ( 2 2 ω 0 ) j ′ = 1 24 2 m a 2 ω 0 i ′ + 1 24 2 m a 2 ω 0 j ′ (2) Calculate the angular velocity at B ( v B ) using the formula: v B = ω × r B / A Here, ω is the angular velocity of the rotating plate and r B / A is the distance of B with respect to A . Substitute ω x ′ i ′ + ω y ′ j ′ + ω z ′ k ′ for ω and − a j ′ for r B / A . v B = ( ω x ′ i ′ + ω y ′ j ′ + ω z ′ k ′ ) × ( − a j ′ ) The matrix multiplication for vector product is done. v B = a ( ω z ′ i ′ + ω x ′ k ′ ) The corner B does not rebound after impact. Therefore the velocity of B after impact in the z ′ axis ( v B ) z ′ is zero. The angular velocity in the x ′ axis ( ω x ′ ) is also zero. Calculate the angular velocity about the mass center ( v ¯ ) using the formula: v ¯ = ω × r G / A Here, r G / A is the distance of mass center with respect to A . Substitute ω x ′ i ′ + ω y ′ j ′ + ω z ′ k ′ for ω and 1 2 a ( − i ′ − j ′ ) for r G / A . v ¯ = ( ω x ′ i ′ + ω y ′ j ′ + ω z ′ k ′ ) × 1 2 a ( − i ′ − j ′ ) Substitute 0 for ω x ′ . v ¯ = ( ω y ′ j ′ + ω z ′ k ′ ) × 1 2 a ( − i ′ − j ′ ) The matrix multiplication for vector product is done. v ¯ = 1 2 a ( ω z ′ i ′ − ω z ′ j ′ + ω y ′ k ′ ) (3) The expression for the angular momentum about A as follows: H A = H G + r G / A × m v ¯ (4) Here, H A is the angular momentum about A and H G is the angular momentum about the mass center. Calculate the angular momentum about G using the formula: H G = I ¯ x ′ ω x ′ i ′ + I ¯ y ′ ω y ′ j ′ + I ¯ z ′ ω z ′ k ′ Substitute 0 for ω x ′ , 1 12 m a 2 for I ¯ y ′ , and 1 6 m a 2 for I ¯ z ′ . H G = ( 1 12 m a 2 ) ω y ′ j ′ + ( 1 6 m a 2 ) ω z ′ k ′ (5) Substitute ( 1 12 m a 2 ) ω y ′ j ′ + ( 1 6 m a 2 ) ω z ′ k ′ for H G , 1 2 a ( − i ′ − j ′ ) for r G / A , and 1 2 a ( ω z ′ i ′ − ω z ′ j ′ − ω y ′ k ′ ) for v ¯ in Equation (4) (b) To determine Find the velocity ( v ¯ ) of its mass center G .

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 18.1, Problem 18.31P

(a)

To determine

The angular velocity (ω) of the plate.

(b)

To determine

Find the velocity (v¯) of its mass center G.

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Chapter 18.1, Problem 18.30P

Chapter 18.1, Problem 18.32P

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The angular velocity ( ω ) of the plate.

Solution Summary: The angular velocity of the plate is underset_18omega0(j-i). The length of a square is obtained by